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Problem 2.85 Determine the direction cosines of the
vectors FAand FB.
y
60⬚
FB
FA
Solution: We have the vectors
FAD140 lb⊲[cos 40°sin 50°]iC[sin 40°]jC[cos 40°cos 50°]k⊳
Problem 2.86 In Example 2.8, suppose that a change
in the wind causes a change in the position of the balloon
and increases the magnitude of the force Fexerted on
the hook at Oto 900 N. In the new position, the angle
y
Problem 2.87 An engineer calculates that the magni-
22.0, 18.4) m and (9.2, 24.4, 15.6) m, respectively.
Express the force Pin terms of scalar components.
Solution: The components of the position vector from Bto Aare
D⊲12.4C9.2⊳iC⊲22.024.4⊳j
C⊲18.4C15.6⊳k
D7.65 eBA
D5.01i3.76j4.39k⊲kN⊳.
Problem 2.88 The cable BC exerts an 8-kN force F
on the bar AB at B.
(a) Determine the components of a unit vector that
y
B (5, 6, 1) m
48
Problem 2.89 A cable extends from point Cto
point E. It exerts a 50-lb force Ton plate Cthat is
directed along the line from Cto E. Express Tin terms
of components.
Dx
A
6 ft
E
y
eCE D⊲04⊳iC⊲2⊲1.37⊳⊳jC⊲63.76⊳k
p42C3.372C2.242
eCE D0.703iC0.592jC0.394k
TD50eCE ⊲lb⊳
TD35.2iC29.6jC19.7k⊲lb⊳
C
B
20
°
4 ft
4 ft
2 ft
z
T
Problem 2.90 In Example 2.9, suppose that the metal
loop at Ais moved upward so that the vertical distance to
Aincreases from 7 ft to 8 ft. As a result, the magnitudes
of the forces FAB and FAC increase to jFABjDjFACjD
240 lb. What is the magnitude of the total force FD
FAB CFAC exerted on the loop by the rope?
Solution: The new coordinates of point Aare (6, 8, 0) ft. The
position vectors are
rAB D⊲4i8jC4k⊳ft
rAC D⊲4i8jC6k⊳ft
Problem 2.91 The cable AB exerts a 200-lb force FAB
at point Athat is directed along the line from Ato B.
Express FAB in terms of components.
C
8 ft
y
8 ft
8.718 iC6
8.718 j2
8.718 k
or
uAB D0.6882iC0.6882j0.2294k.
Problem 2.92 Consider the cables and wall described
in Problem 2.91. Cable AB exerts a 200-lb force FAB
at point Athat is directed along the line from Ato B.
The cable AC exerts a 100-lb force FAC at point Athat
is directed along the line from Ato C. Determine the
50
Problem 2.93 The 70-m-tall tower is supported by
three cables that exert forces FAB,FAC , and FAD on it.
The magnitude of each force is 2 kN. Express the total
force exerted on the tower by the three cables in terms
of components.
A
y
D
FAB
FAC
FAD
A
rAC D40i70jC40k
rAB D⊲40 0⊳iC⊲070⊳jC⊲00⊳k
rAB D40i70jC0k
The unit vectors corresponding to these position vectors are:
uAD DrAD
Problem 2.94 Consider the tower described in Pro-
blem 2.93. The magnitude of the force FAB is 2 kN. The
xand zcomponents of the vector sum of the forces
exerted on the tower by the three cables are zero. What
are the magnitudes of FAC and FAD?
FAB DjFABjuAB D0.9926i1.737jC0k
The forces FAC and FAD are:
FAD DjFADjuAD DjFAD j⊲0.5455i0.6364j0.5455k⊳
0.4444jFACj0.5455jFADjD0
These can be solved by means of standard algorithms, or by the use of
Problem 2.95 In Example 2.10, suppose that the
distance from point Cto the collar Ais increased from
0.2 m to 0.3 m, and the magnitude of the force T
increases to 60 N. Express Tin terms of its components.
0.2 m
z
D
0.4 m
y
C
B
The coordinates of Aare (0.263, 0.0949, 0.171) m.
The position vector from Ato Bis
52
Problem 2.96 The cable AB exerts a 32-lb force Ton
the collar at A. Express Tin terms of components.
y
6 ft
B
4 ft
T
Solution: The coordinates of point Bare B(0, 7, 4). The vector
Problem 2.97 The circular bar has a 4-m radius and
lies in the x-yplane. Express the position vector from
y
Solution: From the figure, the point Bis at (0, 4, 3) m. The coor-
dinates of point Aare determined by the radius of the circular bar
are (3.76, 1.37, 0) m. The vector from Bto Ais given by rBA D⊲xA
xB⊳iC⊲yAyB⊳jC⊲zAzB⊳kD3.76i2.63j3km. Finally, the
Problem 2.98 The cable AB in Problem 2.97 exerts a
60-N force Ton the collar at Athat is directed along the
Solution: We know rBA D3.76i2.63j3km from Problem
2.97. The unit vector uAB DrBA/jrBAj. The unit vector is uAB D
Problem 2.99 In Active Example 2.11, suppose that
the vector Vis changed to VD4i6j10k.
(a) What is the value of UžV?
(b) What is the angle between Uand Vwhen they are
placed tail to tail?
Solution: From Active Example 2.4 we have the expression for U.
Thus
UD6i5j3k,VD4i6k10k
Problem 2.100 In Example 2.12, suppose that the coor-
dinates of point Bare changed to (6, 4, 4) m. What is
the angle between the lines AB and AC?
z
Solution: Using the new coordinates we have
rAB D⊲2iCjC2k⊳m,jrABjD3m
Problem 2.101 What is the dot product of the position
vector rD10iC25j(m) and the force vector
FD300iC250jC300k⊲N⊳?
Solution: Use Eq. (2.23).
FÐrD⊲300⊳⊲10⊳C⊲250⊳⊲25⊳C⊲300⊳⊲0⊳D3250 N-m
Problem 2.102 Suppose that the dot product of two
vectors Uand Vis UÐVD0. If jUj 6D 0, what do you
know about the vector V?
Solution:
Either jVjD0orV?U
54
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.103 Two perpendicular vectors are given
in terms of their components by
Solution: When the vectors are perpendicular, UÐV0.
Thus
UxD8.67
Problem 2.104 Three vectors
UDUxiC3jC2k
Solution: For mutually perpendicular vectors, we have three
equations, i.e.,
UÐVD0
VyD0.857
WzD3.143
20.
(a) Use the definition of the dot product to determine
UÐV.
(b) Use Eq. (2.23) to obtain UÐV.
V
U
Problem 2.106 By evaluating the dot product UÐV,
prove the identity cos⊲12⊳Dcos 1cos 2Csin 1
sin 2.
Strategy: Evaluate the dot product both by using
y
Problem 2.107 Use the dot product to determine the
angle between the forestay (cable AB) and the backstay
(cable BC).
y
B (4, 13) m
Solution: The unit vector from Bto Ais
eBA DrBA
jrBAjD0.321i0.947j
The unit vector from Bto Cis
eBC DrBC
jrBCjD0.385i0.923j
56
Problem 2.108 Determine the angle between the
lines AB and AC (a) by using the law of cosines (see
Appendix A); (b) by using the dot product. (4, 3, ⫺1) m
B
y
A
Solution:
Problem 2.109 The ship Omeasures the positions of
the ship Aand the airplane Band obtains the coordinates
shown. What is the angle between the lines of sight
OA and OB?
y
B
(4, 4, ⫺4) km
Solution: From the coordinates, the position vectors are:
Problem 2.110 Astronauts on the space shuttle use
radar to determine the magnitudes and direction cosines
0.384, cos zD0.512. The vector rBfrom the shuttle
to satellite Bhas magnitude 4 km and direction cosines
cos xD0.743, cos yD0.557, cos zD0.371. What
is the angle between the vectors rAand rB?
x
rB
B
θ
Problem 2.111 In Example 2.13, if you shift your
position and the coordinates of point Awhere you apply
the 50-N force become (8, 3, 3) m, what is the vector
component of Fparallel to the cable OB?
y
(6, 6, –3) m
F
A
z
rOA D⊲8iC3j3k⊳m
58
Problem 2.112 The person exerts a force FD60i
40j(N) on the handle of the exercise machine. Use
Eq. (2.26) to determine the vector component of Fthat
250 mm
200 mm
z
x
Solution: The vector rfrom the Oto where the person grips the
handle is
rD⊲250iC200j150k⊳mm,
C0.566j0.424k⊳
Problem 2.113 At the instant shown, the Harrier’s
thrust vector is TD17,000iC68,000j8,000k(N)
and its velocity vector is vD7.3iC1.8j0.6k(m/s).
The quantity PDjTpjjvj, where Tpis the vector
y
Problem 2.114 Cables extend from Ato Band from
Ato C. The cable AC exerts a 1000-lb force Fat A.
(a) What is the angle between the cables AB and AC?
A(0, 7, 0) ft
B
C
x
y
(0, 0, 10) ft
Solution: Use Eq. (2.24) to solve.
(a) From the coordinates of the points, the position vectors are:
rAB D⊲00⊳iC⊲07⊳jC⊲10 0⊳k
60
Problem 2.115 Consider the cables AB and AC shown
in Problem 2.114. Let rAB be the position vector from
Solution: From Problem 2.114, rAB D0i7jC10k, and eAC D
0.6667i0.3333jC0.6667k. Thus rAB ÐeAC D9, and ⊲rAB ÐeAC⊳eAC
Problem 2.116 The force FD10iC12j6k⊲N⊳.
Determine the vector components of Fparallel and nor-
mal to line OA.
y
(0, 6, 4) m
A
7.21 jC4
7.21 kD0.832jC0.555k
FPD[⊲10iC12j6k⊳Ð⊲0.832jC0.555k⊳]eOA
FPD[6.656]eOA D0iC5.54jC3.69k⊲N⊳
0.4 m
0.5 m
0.3 m
0.2 m
0.25 m
0.2 m
z
x
A
BC
D
O
Solution: We have the following vectors
rCD D⊲0.2i0.3jC0.25k⊳m
jrCDjD⊲0.456i0.684jC0.570k⊳
rOC D⊲0.4iC0.3j⊳m
rOA DrOC C⊲0.2 m⊳eCD D⊲0.309iC0.163jC0.114k⊳m
rAB DrOB rOA D⊲0.309iC0.337jC0.036k⊳m
0.2 m
z
D
TnD⊲37.1iC31.6jC8.22k⊳N
62
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.119 The disk Ais at the midpoint of the
sloped surface. The string from Ato Bexerts a 0.2-lb
force Fon the disk. If you express Fin terms of vector
components parallel and normal to the sloped surface,
what is the component normal to the surface?
y
B
F
(0, 6, 0) ft
The unit vector eAB can be found by
eAB D⊲xBxA⊳iC⊲yByA⊳jC⊲zBzA⊳h
⊲xBxA⊳2C⊲yByA⊳2C⊲zBzA⊳2
Point Bis at (0, 6, 0) (ft) and Ais at (5, 1, 4) (ft).
Substituting, we get
eAB D0.615iC0.615j0.492k
Now FDjFjeAB D⊲0.2⊳eAB
FD0.123iC0.123j0.0984k⊲lb⊳
The component of Fnormal to the surface is the component parallel
to the unit vector eN.
FNORMAL D⊲FÐeN⊳eND⊲0.955⊳eN
FNORMAL D0iC0.0927jC0.0232klb
z8
Problem 2.120 In Problem 2.119, what is the vector
component of Fparallel to the surface?
Problem 2.121 An astronaut in a maneuvering unit
approaches a space station. At the present instant, the
station informs him that his position relative to the origin
of the station’s coordinate system is rGD50iC80jC
y
zx
A
G
Problem 2.122 In Problem 2.121, determine the vec-
tor component of the astronaut’s velocity parallel to the
line from his position to the airlock’s position.
Solution: The coordinates are A(12, 0, 20) m, G(50, 80, 180) m.
Therefore
64
Problem 2.123 Point Pis at longitude 30°W and lati-
tude 45°N on the Atlantic Ocean between Nova
Scotia and France. Point Qis at longitude 60°E and
latitude 20°N in the Arabian Sea. Use the dot product to
y
N
and (2.24), the angular separation of Pand Qis given by
cos DPÐQ
jPjjQj.
PDrOB CrBP DRE⊲icos Pcos PCjsin PCksin Pcos P⊳.
A similar argument for the point Qyields
QDrOC CrCQ DRE⊲icos Qcos QCjsin QCksin Qcos Q⊳
Using the identity cos2ˇCsin2ˇD1, the magnitudes are
jPjDjQjDRE
PÐQDR2
E⊲cos⊲PQ⊳cos Pcos QCsin Psin Q⊳
Substitute:
x
30°
45°
60°
20°
RE
θ
Q
P
G
bc
Problem 2.124 In Active Example 2.14, suppose that
the vector Vis changed to VD4i6j10k.
(a) Determine the cross product UðV. (b) Use the dot
product to prove that UðVis perpendicular to V.
Solution: We have UD6i5jk,VD4k6j10k
(a) UðVD
ij k
651
4610
D44iC56j16k
UðVD44iC56j16k
(b) ⊲UðV⊳ÐVD⊲44⊳⊲4⊳C⊲56⊳⊲6⊳C⊲16⊳⊲10⊳D0)
⊲UðV⊳?V
